A catalyst is known to speed up both forward/backward reactions of a reversible reaction. But how does this work, because the mechanism for the forward and backward reactions could be different right? I mean the adsorbed species for the forward and backward reactions when conducted independently will be different right? Is that because all the steps are exactly reversible?


4 Answers 4


The idea is that the activation energies is lowered for both the forward and backward reactions.

Reaction Energy

Though the activation energy for the backward reaction is higher than the activation energy for the forward reaction, it is nevertheless lowered.

To answer your second question, reactions aren't always reversible, but a lot of the time they are, and are subject to the principle of microscopic reversibility. That means when reactions are performed as a series of steps, at equilibrium, each individual reaction occurs in such a way that the forward and reverse rates of reaction are equal, and these equilibriums can be affected to drive the reaction one way or the other.

  • $\begingroup$ how does this assumption of microscopic reversibility hold good in reality?can I assume a catalyst that accelerates the forward reaction can also be used to catalyze the baxckward reaction always(given thermodynamic limitations of course)?sometimes catalyst chemically decompose in presence of some reatants and it is this decomposed state that actively catalyzes the reaction. But when you start with the products and do the reverse reaction I am not sure if the same exact decomposed states woud be formed $\endgroup$
    – daraj
    Commented Mar 9, 2016 at 7:21
  • $\begingroup$ It's very important in synthetic organic chemistry. It simplifies what you need to know (if you know the forward reaction you know the backward reaction as well), it can help you to predict side products, and it can make undoing a reaction easier by understanding what changes need to be made to affect the equilibrium one way or the other. $\endgroup$
    – ringo
    Commented Mar 9, 2016 at 7:23
  • $\begingroup$ so any catalyst that accelerates forward reaction(in heterogeneous gas-solid reactions) will be equally effective if I need to accelerate the reverse reaction as well? Is this true under all circumstances, just for confirmation? $\endgroup$
    – daraj
    Commented Mar 9, 2016 at 7:38
  • $\begingroup$ Yes. This doesn't change yield at all, but the rates of reaction are effected equally. $\endgroup$
    – ringo
    Commented Mar 9, 2016 at 7:50

The equilibrium constant for the reaction is equal to the ratio of the forward rate constant to the reverse rate constant. In order for the equilibrium constant to be independent of the catalytic effect, the forward and reverse rates must speed up equally.


One of the implications of the second law of thermodynamics is that, if a catalyst increases the rate of the forward reaction by a factor of, say, 745492, it obligatorily increases the rate of the reverse reaction by the same factor, 745492, despite the fact that the two reactions - forward and reverse - may be entirely different (e.g. the diffusion factor is crucial for one but not important for the other) and accordingly require entirely different catalytic strategies. The absurd implication is usually referred to as "Catalysts do not shift chemical equilibrium".

A paradigmatic example of the inability of catalysts to accelerate both - forward and reverse - reactions to the same extent are diffusion-controlled reactions:

Wikipedia: "Diffusion-controlled (or diffusion-limited) reactions are reactions that occur so quickly that the reaction rate is the rate of transport of the reactants through the reaction medium (usually a solution). As quickly as the reactants encounter each other, they react. The process of chemical reaction can be considered as involving the diffusion of reactants until they encounter each other in the right stoichiometry and form an activated complex which can form the product species. The observed rate of chemical reactions is, generally speaking, the rate of the slowest or "rate determining" step. In diffusion controlled reactions the formation of products from the activated complex is much faster than the diffusion of reactants and thus the rate is governed by collision frequency."

Can a catalyst accelerate a diffusion-controlled reaction? Obviously not - catalysts do not accelerate diffusion. On the other hand, nothing prevents the catalyst from accelerating the reverse reaction, which means shifting the equilibrium and is tantamount to violating the second law of thermodynamics:

Consider the association-dissociation reaction

A + B <-> C

which is in equilibrium. Let us assume that the forward reaction

A + B -> C

is diffusion-controlled. The reverse

C -> A + B

is not diffusion-controlled, obviously.

We add a catalyst, e.g. a macroscopic catalytic surface, and it starts splitting C so efficiently that the rate of the reverse (dissociation) reaction increases by a factor of, say, 745492. If the second law of thermodynamics is obeyed, the catalyst must increase the rate of the forward (association) reaction by exactly the same factor, 745492, which is impossible - the catalyst cannot accelerate the forward reaction at all!

Scientists have always known that catalysts do shift chemical equilibirum:

https://www.facebook.com/ParadigmEnergy/posts/249600938581128 "For 50 years scientists have seen in experiments that some monomers and dimers split apart and rejoin at different rates on different surfaces. The eureka moment came when we recognized that by placing two different surfaces close together in a way that effectively eliminates the gas cloud, the energy balance would be different on each of the two surfaces. One surface would have more molecules breaking apart, cooling it, while the other surface would have more molecules joining back together, warming it."

The second-law-violating effect is presented by Wikipedia as a fact:

Wikipedia: "Epicatalysis is a newly identified class of gas-surface heterogeneous catalysis in which specific gas-surface reactions shift gas phase species concentrations away from those normally associated with gas-phase equilibrium. [...] A traditional catalyst adheres to three general principles, namely: 1) it speeds up a chemical reaction; 2) it participates in, but is not consumed by, the reaction; and 3) it does not change the chemical equilibrium of the reaction. Epicatalysts overcome the third principle..."

Wikipedia: "Consider a dimeric gas (A2) that is susceptible to endothermic dissociation or exothermic recombination (A2 <-> 2A). The gas is housed between two surfaces (S1 and S2), whose chemical reactivities are distinct with respect to the gas. Specifically, let S1 preferentially dissociate dimer A2 and desorb monomer A, while S2 preferentially recombines monomers A and desorbs dimer A2. [...]


In 2014 Duncan's temperature paradox was experimentally realized, utilizing hydrogen dissociation on high-temperature transition metals (tungsten and rhenium). Ironically, these experiments support the predictions of the paradox and provide laboratory evidence for second law breakdown."


The point missing so far in the answers is that a catalyst causes the reaction to occur by a different mechanism than it would normally take. (This is why its actually very hard to find good catalysts)

The catalyst lowers the activation energy but its not the same barrier as without the catalyst. So in that sense it does not 'lower the barrier' as commonly stated.

As the reactant and products are still at the same relative energy (no matter if catalysed or not) but the reaction occurs by a lower activation barrier then both forward and reverse rate constants are speeded up.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.